If tan (A - B) = x,then the value of x is | vedclass

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(D) We know the trigonometric identity for the tangent of a difference of two angles:$	an (A - B) = \frac{\sin (A - B)}{\cos (A - B)}$Using the expan

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$\frac{	an A - 	an B}{1 + 	an A 	an B}$

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(D) We know the trigonometric identity for the tangent of a difference of two angles:$	an (A - B) = \frac{\sin (A - B)}{\cos (A - B)}$Using the expansion formulas for sine and cosine of a difference:$\sin (A - B) = \sin A \cos B - \cos A \sin B$$\cos (A - B) = \cos A \cos B + \sin A \sin B$Substituting these into the expression for $	an (A - B)$:$	an (A - B) = \frac{\sin A \cos B - \cos A \sin B}{\cos A \cos B + \sin A \sin B}$Dividing both the numerator and the denominator by $\cos A \cos B$:$	an (A - B) = \frac{\frac{\sin A \cos B}{\cos A \cos B} - \frac{\cos A \sin B}{\cos A \cos B}}{\frac{\cos A \cos B}{\cos A \cos B} + \frac{\sin A \sin B}{\cos A \cos B}}$Simplifying the terms:$	an (A - B) = \frac{	an A - 	an B}{1 + 	an A 	an B}$Thus,$x = \frac{	an A - 	an B}{1 + 	an A 	an B}$.

If $\tan (A - B) = x$,then the value of $x$ is

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